21.(10分)如图,某小区有两个喷泉A,B,两个喷泉的距离为125 m.现要为喷泉铺设供水管道AM,BM,供水点M在小路AC上,供水点M到AB的距离MN的长为60 m,BM的长为75 m.
(1)求供水点M到喷泉A,B需要铺设的管道总长;
(2)求喷泉B到小路AC的最短距离.
(1)求供水点M到喷泉A,B需要铺设的管道总长;
(2)求喷泉B到小路AC的最短距离.
答案：21.解:(1)在$\mathrm{Rt}△MNB$中,
$\because BM=75\ \mathrm{m}$,$MN=60\ \mathrm{m}$,
$\therefore BN=\sqrt{BM^2-MN^2}=\sqrt{75^2-60^2}=45(\mathrm{m})$,
$\therefore AN=AB-BN=125-45=80(\mathrm{m})$,
$\therefore$在$\mathrm{Rt}△AMN$中,$AM=\sqrt{AN^2+MN^2}=\sqrt{80^2+60^2}=100(\mathrm{m})$.
$\therefore AM+BM=100+75=175(\mathrm{m})$,
$\therefore$供水点$M$到喷泉$A,B$需要铺设的管道总长为$175\ \mathrm{m}$.
(2)$\because AB=125\ \mathrm{m}$,$AM=100\ \mathrm{m}$,$BM=75\ \mathrm{m}$,
$\therefore AB^2=BM^2+AM^2$,
$\therefore △ABM$是直角三角形,
$\therefore BM⊥AC$,$\therefore$喷泉$B$到小路$AC$的最短距离是$75\ \mathrm{m}$.
$\because BM=75\ \mathrm{m}$,$MN=60\ \mathrm{m}$,
$\therefore BN=\sqrt{BM^2-MN^2}=\sqrt{75^2-60^2}=45(\mathrm{m})$,
$\therefore AN=AB-BN=125-45=80(\mathrm{m})$,
$\therefore$在$\mathrm{Rt}△AMN$中,$AM=\sqrt{AN^2+MN^2}=\sqrt{80^2+60^2}=100(\mathrm{m})$.
$\therefore AM+BM=100+75=175(\mathrm{m})$,
$\therefore$供水点$M$到喷泉$A,B$需要铺设的管道总长为$175\ \mathrm{m}$.
(2)$\because AB=125\ \mathrm{m}$,$AM=100\ \mathrm{m}$,$BM=75\ \mathrm{m}$,
$\therefore AB^2=BM^2+AM^2$,
$\therefore △ABM$是直角三角形,
$\therefore BM⊥AC$,$\therefore$喷泉$B$到小路$AC$的最短距离是$75\ \mathrm{m}$.
22.(10分)如图,某小区有一块四边形空地ABCD,其中AB⊥BC,AB=9 m,BC=12 m,CD=17 m,AD=8 m,为了美化环境,计划将这块四边形空地进行绿化建设.
(1)求需要绿化的空地ABCD的面积；
(2)为方便居民出入,设计了小路AE,且AE⊥CD于点E,试求小路AE的长.
(1)求需要绿化的空地ABCD的面积；
(2)为方便居民出入,设计了小路AE,且AE⊥CD于点E,试求小路AE的长.
答案：
22.解:如答图,连接$AC$.
(1)$\because AB⊥BC$,$\therefore ∠ABC=90°$.
$\because AB=9\ \mathrm{m}$,$BC=12\ \mathrm{m}$,
$\therefore AC=\sqrt{AB^2+BC^2}=\sqrt{9^2+12^2}=15(\mathrm{m})$.
$\because CD=17\ \mathrm{m}$,$AD=8\ \mathrm{m}$,
$\therefore AD^2+AC^2=8^2+15^2=289=17^2=CD^2$,
$\therefore △DAC$为直角三角形,$∠DAC=90°$,
$\therefore$需要绿化的空地$ABCD$的面积为$S_{△ABC}+S_{△DAC}=\dfrac{1}{2}AB·BC+\dfrac{1}{2}AD·AC=\dfrac{1}{2}×9×12+\dfrac{1}{2}×8×15=114(\mathrm{m}^2)$.
(2)$\because AE⊥CD$,$∠DAC=90°$,
$\therefore S_{△DAC}=\dfrac{1}{2}AD·AC=\dfrac{1}{2}CD·AE$,
$\therefore 8×15=17×AE$,解得$AE=\dfrac{120}{17}(\mathrm{m})$.
22.解:如答图,连接$AC$.
(1)$\because AB⊥BC$,$\therefore ∠ABC=90°$.
$\because AB=9\ \mathrm{m}$,$BC=12\ \mathrm{m}$,
$\therefore AC=\sqrt{AB^2+BC^2}=\sqrt{9^2+12^2}=15(\mathrm{m})$.
$\because CD=17\ \mathrm{m}$,$AD=8\ \mathrm{m}$,
$\therefore AD^2+AC^2=8^2+15^2=289=17^2=CD^2$,
$\therefore △DAC$为直角三角形,$∠DAC=90°$,
$\therefore$需要绿化的空地$ABCD$的面积为$S_{△ABC}+S_{△DAC}=\dfrac{1}{2}AB·BC+\dfrac{1}{2}AD·AC=\dfrac{1}{2}×9×12+\dfrac{1}{2}×8×15=114(\mathrm{m}^2)$.
(2)$\because AE⊥CD$,$∠DAC=90°$,
$\therefore S_{△DAC}=\dfrac{1}{2}AD·AC=\dfrac{1}{2}CD·AE$,
$\therefore 8×15=17×AE$,解得$AE=\dfrac{120}{17}(\mathrm{m})$.
23.(12分)在$△ ABC$中，$∠ ACB=90°$，D为$△ ABC$内一点，连接BD，DC，延长DC到点E，使得$CE=DC$.
(1)如图①，延长BC到点F，使得$CF=BC$，连接AF，EF.若$AF⊥ EF$，求证：$BD⊥ AF$；
(2)连接AE，交BD的延长线于点H，连接CH，依题意补全图②.若$AB^2=AE^2+BD^2$，用等式表示线段CD与CH之间的数量关系，并说明理由.
(1)如图①，延长BC到点F，使得$CF=BC$，连接AF，EF.若$AF⊥ EF$，求证：$BD⊥ AF$；
(2)连接AE，交BD的延长线于点H，连接CH，依题意补全图②.若$AB^2=AE^2+BD^2$，用等式表示线段CD与CH之间的数量关系，并说明理由.
答案：
23.(1)证明:如答图①,延长$BD$交$AF$于点$M$.
$\because AF⊥EF$,$\therefore ∠AFE=90°$.
在$△BCD$和$△FCE$中,$\begin{cases} BC=FC, \\ ∠BCD=∠FCE, \\ DC=EC, \end{cases}$
$\therefore △BCD≌△FCE(\mathrm{SAS})$,$\therefore ∠CBD=∠CFE$,
$\therefore BD// EF$,$\therefore ∠AMB=∠AFE=90°$,$\therefore BD⊥AF$.
(2)解:补全图形如答图②,$CH=CD$.
理由如下:
如答图②,延长$BC$至点$F$,使$CF=BC$,连接$AF,EF$.
$\because ∠ACB=90°$,$\therefore AC⊥BF$,$\therefore AB=AF$.
由(1)可知$BH// EF$,$BD=EF$.
$\because AB^2=AE^2+BD^2$,$\therefore AF^2=AE^2+EF^2$,
$\therefore ∠AEF=90°$.
$\because BH// EF$,$\therefore ∠BHE=90°$.
又$\because CE=DC$,$\therefore CH=\dfrac{1}{2}DE=CD$.
23.(1)证明:如答图①,延长$BD$交$AF$于点$M$.
$\because AF⊥EF$,$\therefore ∠AFE=90°$.
在$△BCD$和$△FCE$中,$\begin{cases} BC=FC, \\ ∠BCD=∠FCE, \\ DC=EC, \end{cases}$
$\therefore △BCD≌△FCE(\mathrm{SAS})$,$\therefore ∠CBD=∠CFE$,
$\therefore BD// EF$,$\therefore ∠AMB=∠AFE=90°$,$\therefore BD⊥AF$.
(2)解:补全图形如答图②,$CH=CD$.
理由如下:
如答图②,延长$BC$至点$F$,使$CF=BC$,连接$AF,EF$.
$\because ∠ACB=90°$,$\therefore AC⊥BF$,$\therefore AB=AF$.
由(1)可知$BH// EF$,$BD=EF$.
$\because AB^2=AE^2+BD^2$,$\therefore AF^2=AE^2+EF^2$,
$\therefore ∠AEF=90°$.
$\because BH// EF$,$\therefore ∠BHE=90°$.
又$\because CE=DC$,$\therefore CH=\dfrac{1}{2}DE=CD$.